A small amount, 6.78 g, of pentane (C₅H₁₂) is mixed with 100.00 g hexane (C₆H₁₄). Calculate the vapour pressure of the resulting solution. The vapour pressures of pure pentane and hexane are 425 Torr and 151 Torr, respectively. Assume ideal behaviour. A) 201 Torr B) 158 Torr C) 183 Torr D) 164 Torr E) 172 Torr

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The vapor pressure of the solution is calculated using Raoult's Law: P_solution = (X_pentane * P_pentane) + (X_hexane * P_hexane). First, calculate the moles of pentane and hexane, then their mole fractions. Moles of pentane = 6.78 g / 72.15 g/mol = 0.094 moles. Moles of hexane = 100.00 g / 86.18 g/mol = 1.16 moles. Total moles = 0.094 + 1.16 = 1.254 moles. Mole fraction of pentane = 0.094 / 1.254 = 0.075. Mole fraction of hexane = 1.16 / 1.254 = 0.925. P_solution = (0.075 * 425 Torr) + (0.925 * 151 Torr) = 31.875 + 139.675 = 171.55 Torr, approximately 172 Torr.

A Small Amount, 6